Divide the following complex numbers. $ \dfrac{9-i}{-4+5i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4-5i}$ $ \dfrac{9-i}{-4+5i} = \dfrac{9-i}{-4+5i} \cdot \dfrac{{-4-5i}}{{-4-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(9-i) \cdot (-4-5i)} {(-4+5i) \cdot (-4-5i)} = \dfrac{(9-i) \cdot (-4-5i)} {(-4)^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(9-i) \cdot (-4-5i)} {(-4)^2 - (5i)^2} = $ $ \dfrac{(9-i) \cdot (-4-5i)} {16 + 25} = $ $ \dfrac{(9-i) \cdot (-4-5i)} {41} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({9-i}) \cdot ({-4-5i})} {41} = $ $ \dfrac{{9} \cdot {(-4)} + {-1} \cdot {(-4) i} + {9} \cdot {-5 i} + {-1} \cdot {-5 i^2}} {41} $ Evaluate each product of two numbers. $ \dfrac{-36 + 4i - 45i + 5 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{-36 + 4i - 45i - 5} {41} = \dfrac{-41 - 41i} {41} = -1-i $